Proof.
The interlacing of the eigenvalues of \(A\)  and \(J\) , i.e., the way \(h(\lambda )\)  and \(g(\lambda )\)  intersect, follows a similar result by Lemma 3.3 in [1]}. Because of the fact that \(A\)  is nonderogatory, in cases 1b, 3b, 4b and 4c there is a Jordan block of size two corresponding to the two eigenvalues that coincide, while in case 4d there is a Jordan block of size three corresponding to the three eigenvalues that coincide. See Section 5.6 in [2]}.
